Answer:
a) x 0 1 2 3 4 5 6 7 8 9 10 11 12 13
p(x) .002 .013 .039 .081 .125 .155 .160 .142 .110 .076 0.047 .026 .014 .007
b) Attached
c) Mean = 6.2
Standard deviation = 2.5
The plor is attached.
d) More than 75 customers per hour is very unlikely.
Step-by-step explanation:
The table for the probabilities is:
x 0 1 2 3 4 5 6 7 8 9 10 11 12 13
p(x) .002 .013 - .081 .125 .155 - .142 .110 .076 - .026 .014 .007
"x" represents 10-minute intervals and P(x) is:
[tex]P(x)=\lambda^x\cdot e^{-\lambda}/x!=6.2^x\cdot e^{-6.2}/x![/tex]
The missing spots are for x=2, x=6 and x=10.
[tex]P(2)=6.2^{2} \cdot e^{-6.2}/2!=38*0.002/2=0.039\\\\P(6)=6.2^{6} \cdot e^{-6.2}/6!=56800*0.002/720=0.160\\\\P(10)=6.2^{10} \cdot e^{-6.2}/10!=83929937*0.002/3628800=0.047\\\\[/tex]
b) Attached
c) The mean and standard deviation of the Poisson distribution are:
[tex]E(x)=\lambda =6.2\\\\\sigma=\sqrt{\lambda}=\sqrt{6.2}\approx2.5[/tex]
Plot attached: (μ) +/- (σ) in blue, (μ) +/- 2(σ) in yelow, and (μ) +/- 3(σ) in green
d) The claim of more than 75 customers per hour means 12.5 customers per 10-minute period.
This value represents 2.52 standards deviation from the mean value (6.2 customers) and has a probability of less than P=0.025.
So we can conclude that more than 75 customers per hour is very unlikely.
